## University Physics with Modern Physics (14th Edition)

(a) The work done on the sled will be equal to the elastic potential energy stored in the spring. $W = \frac{1}{2}mv^2 = \frac{1}{2}kx^2$ $v^2 = \frac{kx^2}{m}$ $v = x~\sqrt{\frac{k}{m}}$ $v = (0.375~m)~\sqrt{\frac{4000~N/m}{70.0~kg}}$ $v = 2.83~m/s$ (b) The work done on the sled will be equal to the magnitude of the change in the elastic potential energy stored in the spring. $W = \frac{1}{2}mv^2 = \frac{1}{2}kx_1^2 - \frac{1}{2}kx_2^2$ $v^2 = \frac{k(x_1^2-x_2^2)}{m}$ $v = \sqrt{\frac{k(x_1^2-x_2^2)}{m}}$ $v = \sqrt{\frac{(4000~N/m)\times ((0.375~m)^2-(0.200~m)^2)}{70.0~kg}}$ $v = 2.40~m/s$